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Math Help - Probability

  1. #1
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    Probability

    An instructor gives her class a set of 10 porblems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the the problems, what is the probability that the student will answer correctly
    a) all 5 problems?
    b) at least 4 of the problems?
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  2. #2
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    Quote Originally Posted by r7iris View Post
    An instructor gives her class a set of 10 porblems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the the problems, what is the probability that the student will answer correctly
    a) all 5 problems?
    b) at least 4 of the problems?
    Use the formula {n\choose m}p^m (1 - p)^{n-m}.
    Where p=7/10 = .7 because that is the probability that she knows.
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  3. #3
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    Hello, r7iris!

    I think I've worked it out . . .


    An instructor gives her class a set of 10 problems with the information
    that the final exam will consist of a random selection of 5 of them.

    If a student has figured out how to do 7 of the the problems,
    what is the probability that the student will answer correctly
    (a) all 5 problems?
    (b) at least 4 of the problems?

    (a) It doesn't matter which five problems the instructor picks.
    We are concerned with whether the student's seven prepared solutions
    . . include the five problems.

    There are: . {10\choose7} \,=\,120 ways that the student can choose 7 problems.

    To get all 5 problems correct, his 7 preparations must include:
    . . the 5 problems chosen by the instructor (one way),
    . . and any 2 of the other 5 problems: . {5\choose2} \,=\,10 ways.
    Hence, he has: . 1\cdot10 \:=\:10 ways to get all 5 problems correct.

    Therefore: . P(\text{5 correct}) \;=\;\frac{10}{120} \;=\;\frac{1}{12}


    (b) To get 4 or 5 problems correct . . .

    We already know there are 10 ways to get 5 correct.

    To get 4 correct, he must choose:
    . . 4 of the chosen 5 problems: . {5\choose4} \,=\,5 ways.
    . . and 3 from the other five: . {5\choose3} \,=\,10 ways.
    Hence, he has: . 5\cdot10 \,=\,50 ways to get 4 correct.


    Therefore, he has: . 10 + 50 \:=\:60 ways to get 4 or 5 correct
    . . and: . P(\text{4 or 5 correct}) \;=\;\frac{60}{120} \:=\:\frac{1}{2}

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